GATE Electrical Engineering 2021

Question 1
Let p and q be real numbers such that p^{2}+q^{2}=1. The eigenvalues of the matrix \begin{bmatrix} p & q\\ q& -p \end{bmatrix} are
A
1 and 1
B
1 and -1
C
j and -j
D
pq and -pq
   
Question 1 Explanation: 
Characteristic equation of A
\begin{aligned} \left|A_{2 \times 2}-\lambda I\right|&=(-1)^{2} \lambda^{2}+(-1)^{1} \text{Tr}(A) \lambda+|A|=0 \\ \lambda^{2}-(p-p) \lambda+\left(-p^{2}-q^{2}\right) &=0 \\ \Rightarrow \qquad \qquad\lambda^{2}-1 &=0 \\ \Rightarrow \qquad\qquad \lambda &=\pm 1 \end{aligned}
Question 2
Let p\left ( z\right )=z^{3}+\left ( 1+j \right )z^{2}+\left ( 2+j \right )z+3, where z is a complex number.
Which one of the following is true?
A
\text{conjugate}\:\left \{ p\left ( z \right ) \right \}=p\left ( \text{conjugate} \left \{ z \right \} \right ) for all z
B
The sum of the roots of p\left ( z \right )=0 is a real number
C
The complex roots of the equation p\left ( z \right )=0 come in conjugate pairs
D
All the roots cannot be real
   
Question 2 Explanation: 
Since sum of the roots is a complex number
\Rightarrow absent one root is complex
So all the roots cannot be real.
Question 3
Let f\left ( x \right ) be a real-valued function such that {f}'\left ( x_{0} \right )=0 for some x _{0} \in\left ( 0,1 \right ), and {f}''\left ( x \right )> 0 for all x \in \left ( 0,1 \right ). Then f\left ( x \right ) has
A
no local minimum in (0,1)
B
one local maximum in (0,1)
C
exactly one local minimum in (0,1)
D
two distinct local minima in (0,1)
   
Question 3 Explanation: 
x_{0} \in(0,1), where f(x)=0 is stationary point
and f^{\prime \prime}(x)>0 \qquad \qquad \forall x \in(0,1)
So \qquad \qquad f^{\prime}\left(x_{0}\right)=0
and \qquad \qquad f^{\prime}(0)>0, \text { where } x_{0} \in(0,1)
Hence, f(x) has exactly one local minima in (0,1)
Question 4
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

A
\text{10 V} in series with 12\:\Omega
B
\text{65 V} in series with 15\:\Omega
C
\text{50 V} in series with 2\:\Omega
D
\text{35 V} in series with 2\:\Omega
   
Question 4 Explanation: 
Given circuit can be resolved as shown below,


V_{T H}=15+50=65 \mathrm{~V}


\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
Question 5
Which one of the following vector functions represents a magnetic field \overrightarrow{B}?
(\hat{X}, \hat{Y} and \hat{Z} are unit vectors along x-axis, y-axis, and z-axis, respectively)
A
10x\hat{X}+20y\hat{Y}-30z\hat{Z}
B
10y\hat{X}+20x\hat{Y}-10z\hat{Z}
C
10z\hat{X}+20y\hat{Y}-30x\hat{Z}
D
10x\hat{X}-30z\hat{Y}+20y\hat{Z}
   
Question 5 Explanation: 
If \vec{B} is magnetic flux density then \vec{\nabla} \cdot \vec{B}=0
\begin{aligned} &\vec{\nabla} \cdot \vec{B}=\frac{\partial B x}{\partial x}+\frac{\partial B y}{\partial y}+\frac{\partial B z}{\partial z}\\ &\frac{\partial}{\partial x}(10 x)+\frac{\partial}{\partial y}(20 y)+\frac{\partial}{\partial z}(-30 z)=\vec{\nabla} \cdot \vec{B}\\ &\qquad \qquad \vec{\nabla} \cdot \vec{B}=10+20-30=0 \end{aligned}
Question 6
If the input x(t) and output y(t) of a system are related as y\left ( t \right )=\text{max}\left ( 0,x\left ( t \right ) \right ), then the system is
A
linear and time-variant
B
linear and time-invariant
C
non-linear and time-variant
D
non-linear and time-invariant
   
Question 6 Explanation: 
\begin{aligned} y(t) &=\max (0, x(t)) \\ &=\left\{\begin{array}{cl} 0, & x(t) \lt 0 \\ x(t), & x(t)\gt 0 \end{array}\right. \end{aligned}


Linearity check:
at input x_{1}(t)=-2, output y_{1}(t)=0
at input x_{2}(t)=1, output y_{2}(t)=1


\therefore system is non-linear because it violates law of additivity.
Check for time-invariance :
Delayed \mathrm{O} / \mathrm{P}:
y\left(t-t_{0}\right)=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right) \gt 0 \\ 0 & x\left(t-t_{0}\right)\lt 0 \end{array}\right.
\mathrm{O} / \mathrm{P} of system when input is x\left(t-t_{0}\right)=f(t)
y_{1}(t)=\left\{\begin{array}{cl} f(t), & f(t) \gt 0 \\ 0, & f(t)\lt 0 \end{array}=\left\{\begin{array}{cl} x\left(t-t_{0}\right), & x\left(t-t_{0}\right)\gt 0 \\ 0, & x\left(t-t_{0}\right)\lt 0 \end{array}\right.\right.
Therefore, system is time-invariant.
Question 7
Two discrete-time linear time-invariant systems with impulse responses h_{1}\left [ n \right ]=\delta \left [ n-1 \right ]+\delta \left [ n+1 \right ] and h_{2}\left [ n \right ]=\delta \left [ n\right ]+\delta \left [ n-1 \right ] are connected in cascade, where \delta \left [ n\right ] is the Kronecker delta. The impulse response of the cascaded system is
A
\delta \left [ n-2\right ]+\delta \left [ n+1 \right ]
B
\delta \left [ n-1\right ]\delta \left [ n\right ]+\delta \left [ n+1 \right ]\delta \left [ n-1 \right ]
C
\delta \left [ n-2\right ]+\delta \left [ n-1\right ]+\delta \left [ n\right ]+\delta \left [ n+1 \right ]
D
\delta \left [ n\right ]\delta \left [ n-1\right ]+\delta \left [ n-2\right ]\delta \left [ n+1 \right ]
   
Question 7 Explanation: 
\begin{aligned} h(n) &=\text { Resultant impulse response } \\ &=h_{1}(n) * h_{2}(n) \end{aligned}
By applying z -transform
\begin{aligned} H(z) &=H_{1}(z) \cdot H_{2}(z) \\ &=\left(z+z^{-1}\right)\left(1+z^{-1}\right) \\ &=z+Z^{-1}+1+z^{2} \end{aligned}
By applying inverse ZT,
h(n)=\delta(n+1)+\delta(n-1)+\delta(n)+\delta(n-2)
Question 8
Consider the table given:
\begin{array}{|c|c|c|} \hline \text{Constructional feature} & \text{Machine type} & \text{Mitigation} \\ \hline \text{P. Damper bars} & \text{S. Induction motor} & \text{X. Hunting} \\ \hline \text{Q. Skewed rotor slots} & \text{T. Transformer} & \text{Y. Magnetic locking} \\ \hline \text{R. Compensating winding} & \text{U. Synchronous machine} & \text{Z. Armature reaction} \\ \hline \text{} & \text{V. DC machine} & \text{} \\ \hline \end{array}
The correct combination that relates the constructional feature, machine type and mitigation is
A
P-V-X, Q-U-Z, R-T-Y
B
P-U-X, Q-S-Y, R-V-Z
C
P-T-Y, Q-V-Z, R-S-X
D
P-U-X, Q-V-Y, R-T-Z
   
Question 8 Explanation: 
P: Damper bars used in synchronous machine (U) to prevent hunting (X)
Q: Skewed rotor slots used in induction motor (S) t avoid magnetic locking (Y)
R: Compensating winding used in DC machine (V) to neutralize cross magnetizing effects of armature reaction (Z) under main poles (polar zone).
Question 9
Consider a power system consisting of N number of buses. Buses in this power system are categorized into slack bus, PV buses and PQ buses for load flow study. The number of PQ buses is N_{L}. The balanced Newton-Raphson method is used to carry out load flow study in polar form. \text{H, S, M, and R} are sub-matrices of the Jacobian matrix J as shown below:
\begin{bmatrix} \Delta P\\ \Delta Q \end{bmatrix}=J\begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}, \text{where}\: J=\begin{bmatrix} H & S\\ M &R \end{bmatrix}
The dimension of the sub-matrix M is
A
N_{L}\times \left ( N-1 \right )
B
\left ( N-1 \right )\times \left ( N-1-N_{L} \right )
C
N_{L}\times \left ( N-1+N_{L} \right )
D
\left ( N-1 \right )\times \left ( N-1+N_{L} \right )
   
Question 9 Explanation: 
\left[\begin{array}{l} \Delta P \\ \Delta Q \end{array}\right]=J\left[\begin{array}{l} \Delta \delta \\ \Delta V \end{array}\right] \text { where } J=\left[\begin{array}{ll} H & S \\ M & R \end{array}\right]
For size of M
Row = No. of unknown variables of Q=N_{L}
Column = No. of variable which has \delta=N_{L}+\left(N-1-N_{L}\right)
\begin{aligned} &=N-1\\ \text{So}, \quad\text{ size of }M&=N_{L} \times(N-1) \end{aligned}
Question 10
Two generators have cost functions F_{1} and F_{2}. Their incremental-cost characteristics are
\dfrac{dF_{1}}{dP_{1}}=40+0.2P_{1}
\dfrac{dF_{2}}{dP_{2}}=32+0.4P_{2}
They need to deliver a combined load of 260 \text{~MW}. Ignoring the network losses, for economic operation, the generations P_{1} and P_{2} (in \text{MW}) are
A
P_{1}=P_{2}=130
B
P_{1}=160, P_{2}=100
C
P_{1}=140, P_{2}=120
D
P_{1}=120, P_{2}=140
   
Question 10 Explanation: 
\begin{aligned} I C_{1} &=I C_{2} \\ 40+0.2 P_{1} &=32+0.4 P_{2} \\ 0.4 P_{2}-0.2 P_{1} &=8 \\ P_{2}+P_{1} &=260 \end{aligned}
Solving equation (i) and (ii),
P_{1}=160 \mathrm{MW} ; P_{2}=100 \mathrm{MW}
There are 10 questions to complete.

GATE Notes – Electrical Engineering

GATE Electrical Engineering notes for all subjects as per syllabus of GATE 2022 Electrical Engineering.

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Power Electronics Miscellaneous

Question 1
A double pulse measurement for an inductively loaded circuit controlled by the IGBT switch is carried out to evaluate the reverse recovery characteristics of the diode. D, represented approximately as a piecewise linear plot of current vs time at diode turn-off. L_{par} is a parasitic inductance due to the wiring of the circuit, and is in series with the diode. The point on the plot (indicate your choice by entering 1. 2, 3 or 4) at which the IGBT experiences the highest current stress is ______.
A
1
B
2
C
3
D
4
GATE EE 2020   Power Electronics
Question 1 Explanation: 


Using KCL, I_s=I_L-I_D
For inductively loaded circuits, load can be assumed to be constant.
\therefore \; I_s is maximum when, I_D is minimum, i.e. at point 3.
Therefore, IGBT experiences highest current stress at point 3.
Question 2
For the circuit shown in the figure below, assume that diodes D_1,D_2 and D_3 are ideal.

The DC components of voltages v_1 \; and \; v_2, respectively are
A
0 V and 1 V
B
-0.5 V and 0.5 V
C
1 V and 0.5 V
D
1 V and 1 V
GATE EE 2017-SET-1   Power Electronics
Question 2 Explanation: 


\begin{aligned} V_{2\; avg}&=\frac{V_m}{\pi}=\frac{\pi/2}{\pi}=\frac{1}{2}=0.5V \\ V_{1\; avg} &=\frac{1}{2 \pi}[ \int_{0}^{\pi}\frac{\pi}{2}\sin 100 \pi t\cdot d(\omega t)\\ &+\int_{\pi}^{2\pi} \pi \sin 100 \pi t\cdot d(\omega t) ]\\ &= -0.5V \end{aligned}
Question 3
Consider a HVDC link which uses thyristor based line-commutated converters as shown in the figure. For a power flow of 750 MW from System 1 to System 2, the voltages at the two ends, and the current, are given by: V_{1}=500 kV, V_{2}=485 kV and V_{3}=1.5 kA. If the direction of power flow is to be reversed (that is, from System 2 to System 1) without changing the electrical connections, then which one of the following combinations is feasible?
A
V_{1}=-500 kV,V_{2}=-485 kV and latex]I=1.5kA[/latex]
B
V_{1}=-485 kV,V_{2}=500 kV and latex]I=1.5kA[/latex]
C
V_{1}=500 kV,V_{2}=-485 kV and latex]I=-1.5kA[/latex]
D
V_{1}=-500 kV,V_{2}=-485 kV and latex]I=-1.5kA[/latex]
GATE EE 2015-SET-1   Power Electronics
Question 3 Explanation: 
To maintain the direction of power flow from system 2 to system 1, the voltage V_1=-485kV and voltage V_2=500kV and I=1.5kA.
Since, current cannot flow in reverse direction. Option (B) is correct answer.
Question 4
The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 \mus is applied to the SCR. The maximum value of R in \Omega to ensure successful firing of the SCR is ______.
A
4050
B
5560
C
6060
D
8015
GATE EE 2014-SET-2   Power Electronics
Question 4 Explanation: 
Let us assume the SCR is conducting,

\begin{aligned} &I_{ss}=\frac{100}{500}=0.2A\\ &[\because \text{inductor will be dhort circuited in DC}]\\ &i(t)=I_{ss}(1-e^{-t/\tau })\\ &\tau =\frac{L}{R}=\frac{200 \times 10^{-3}}{500}\\ &\;\;=4 \times 10^{-4}\; sec\\ &\text{Given }t=50 \times 10^{-6}\; sec\\ &\therefore \; i(t)=0.2\left ( 1-e^{\frac{50 \times 10^{-6}}{4 \times 10^{-4}}} \right )=23.5A \end{aligned}

V=I \times R
R=\frac{V}{I}=\frac{100}{16.5 \times 10^{-3}} =6060 \Omega
Question 5
A single-phase SCR based ac regulator is feeding power to a load consisting of 5 \Omega resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firing angle at which the voltage across the device becomes zero all throughout and the rms value of current through SCR, under this operating condition, are
A
30^{\circ} and 46 A
B
30^{\circ} and 23 A
C
45^{\circ} and 23 A
D
45^{\circ} and 32 A
GATE EE 2014-SET-2   Power Electronics
Question 5 Explanation: 
The maximum firing angle at which the voltage across the device becomes '\phi ' = load angle.
\begin{aligned} \phi &= \tan ^{-1} \left ( \frac{\omega L}{R} \right )\\ &= \tan^{-1}\left ( \frac{2 \pi \times 50 \times 16 \times 10^{-3}}{5} \right ) \\ \phi &=45.15\simeq 45^{\circ} \end{aligned}
Rms value of current through SCR is
\begin{aligned} I_{T_{rms}} &=\sqrt{\left [ \frac{1}{2 \pi}\int_{\alpha }^{\pi+\alpha }\left ( \frac{V_m}{z}\sin (\omega t-\phi ) \right )^2 d(\omega t) \right ]}\\ &=\sqrt{\frac{V_m^2}{2 \pi z^2} \int_{\alpha }^{\pi+\alpha } \left [ \frac{1-\cos 2(\omega t-\phi )}{2} \right ] d(\omega t)}\\ &=\sqrt{\frac{V_m^2}{2 \times 2 \pi z^2} \left [\pi- \left.\begin{matrix} \frac{\sin 2(\omega t-\phi )}{2} \end{matrix}\right|_\alpha ^{(\pi+\alpha )} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} } \sqrt{\left [ \pi+\frac{-\sin 2(\pi+\alpha -\alpha )+\sin 2(\alpha -\alpha )}{2} \right ] }\\ &=\frac{V_m}{2z \sqrt{\pi} }\sqrt{\pi}=\frac{V_m}{2z}\\ I_{T_{rms}}&=\frac{230\sqrt{2}}{2 \times \sqrt{5^2(2 \pi \times 50 \times 16 \times 10^{-3})^2}}\\ &=22.93\simeq 23A\\ I_{T_{rms}}&=23A \end{aligned}
Question 6
A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10\Omega.

The kVA rating of the input transformer is
A
53.2 kVA
B
46.0 kVA
C
22.6 kVA
D
7.5 kVA
GATE EE 2011   Power Electronics
Question 6 Explanation: 
RMS value of supply current in case of 3-\phi bridge converter
I_s=I_0\sqrt{\frac{2}{3}}=40\sqrt{\frac{2}{3}}=32.66A
KVA rating of the input transformrer
\begin{aligned} &=\sqrt{3}V_sI_s \\ &= \sqrt{3} \times 400 \times 32.66 \times 10^{-3}\; kVA\\ &=22.62\; kVA \end{aligned}
Question 7
A solar energy installation utilize a three-phase bridge converter to feed energy into power system through a transformer of 400V/400 V, as shown below.

The energy is collected in a bank of 400 V battery and is connected to converter through a large filter choke of resistance10\Omega.

The maximum current through the battery will be
A
14A
B
40A
C
80A
D
94A
GATE EE 2011   Power Electronics
Question 7 Explanation: 


Average output voltage of the converter
V_0=\frac{3 V_{ml}}{\pi} \cos \alpha
The converter acts as line commutated inverter and for such mode \alpha \gt 90^{\circ} and V_0 is negative.THerefore, battery supplies energy to AC system. So, current through battery
I_0=\frac{400-V_0}{R}
For V_0=0 \text{ or }\alpha =90^{\circ},
Maximum current flow through battery
(I_0)_{max}=\frac{400}{10}=40A
Question 8
The input voltage given to a converter is
v_i=100\sqrt{2}\sin (100 \pi t) V
The current drawn by the converter is
i_i=10\sqrt{2}\sin (100\pi t-\pi/3) + 5\sqrt{2}\sin (300\pi t+\pi/4) + 2\sqrt{2}\sin (500\pi t-\pi/6)A
The active power drawn by the converter is
A
181W
B
500W
C
707W
D
887W
GATE EE 2011   Power Electronics
Question 8 Explanation: 
Rms value of input voltag,
V_{rms}=\frac{100\sqrt{2}}{\sqrt{2}}=100V
Rms value of current,
I_{rms}=\sqrt{\left (\frac{10\sqrt{2}}{\sqrt{2}} \right )^2+\left (\frac{5\sqrt{2}}{\sqrt{2}} \right )^2+\left (\frac{2\sqrt{2}}{\sqrt{2}} \right )^2}=11.358A
Let input power factor \cos \phi
V_{rms}I_{rms} \cos \phi = active power drawn by the coverter
\Rightarrow \;100 \times 11.358 \times \cos \phi =500W
\Rightarrow \; \cos \phi =0.44
Question 9
The input voltage given to a converter is
v_i=100\sqrt{2}\sin (100 \pi t) V
The current drawn by the converter is
i_i=10\sqrt{2}\sin (100\pi t-\pi/3) + 5\sqrt{2}\sin (300\pi t+\pi/4) + 2\sqrt{2}\sin (500\pi t-\pi/6)A
The input power factor of the converter is
A
0.31
B
0.44
C
0.5
D
0.71
GATE EE 2011   Power Electronics
Question 9 Explanation: 
\begin{aligned} V_i&=100\sqrt{2}\sin (100 \pi t)\\ i_i&=10\sqrt{2} \sin \left ( 100 \pi t -\frac{\pi}{3} \right )\\ &+5\sqrt{2} \sin \left ( 300 \pi t +\frac{\pi}{4} \right )\\ &+2\sqrt{2} \sin \left ( 500 \pi t -\frac{\pi}{4} \right )A \end{aligned}
Fundamental component of input voltage
\begin{aligned} (V_i)_1&=100\sqrt{2} \sin (100 \pi t)\\ (V_i)_{1,rms}&=\frac{100\sqrt{2}}{\sqrt{2}}=100V \end{aligned}
Fundamental component of current
\begin{aligned} (i_L)_1&=10\sqrt{2} \sin (100 \pi t-\frac{\pi}{3})\\ (i_L)_{1,rms}&=\frac{10\sqrt{2}}{\sqrt{2}}=10 \end{aligned}
Phase difference between these two components
\phi _1=\frac{\pi}{3}, \cos \phi _1= \cos \frac{\pi}{3}=0.5
Active power due to fundamental components
P_1=(V_i)_{1,rms} \times (i_i)_{1,rms} \cos \phi =100 \times 10 \times 0.5=500W
Since 3^{rd} \text{ and } 5^{th} harmonic are absent in input voltage, there is no active power due to the these components.
Hence, active power drawn by the converter
P_0= Active power due to fundamental components =500 W
Question 10
The power electronic converter shown in the figure has a single-pole double-throw switch. The pole P of the switch is connected alternately to throws A and B. The converter shown is a
A
step down chopper (buck converter)
B
half-wave rectifier
C
step-up chopper (boost converter)
D
full-wave rectifier
GATE EE 2010   Power Electronics
Question 10 Explanation: 
When switch is connected to A for time duration T_1

V_{out}=V_{in}
When switch is connected to B for time duration T_2

Average output voltage =\frac{V_{in}T_1}{T_1+T_2}=\alpha V_{in}
where, \alpha =\text{duty cycle}=\frac{T_1}{T_1+T_2}
Therefore, the converter shown is a step down chooper.
There are 10 questions to complete.

GATE Electrical Engineering-Topic wise Previous Year Questions

GATE 2022 Electrical Engineering Syllabus

Revised syllabus of GATE 2022 Electrical Engineering by IIT.

Practice GATE Electrical Engineering previous year questions

Year wise | Subject wise | Topic wise

Section 1: Engineering Mathematics

Linear Algebra: Matrix Algebra, Systems of linear equations, Eigenvalues, Eigenvectors.
Calculus: Mean value theorems, Theorems of integral calculus, Evaluation of definite and improper integrals, Partial Derivatives, Maxima and minima, Multiple integrals, Fourier series, Vector identities, Directional derivatives, Line integral, Surface integral, Volume integral, Stokes’s theorem, Gauss’s theorem, Divergence theorem, Green’s theorem.
Differential equations: First order equations (linear and nonlinear), Higher order linear differential equations with constant coefficients, Method of variation of parameters, Cauchy’s equation, Euler’s equation, Initial and boundary value problems, Partial Differential Equations, Method of separation of variables.
Complex variables: Analytic functions, Cauchy’s integral theorem, Cauchy’s integral formula, Taylor series, Laurent series, Residue theorem, Solution integrals.
Probability and Statistics: Sampling theorems, Conditional probability, Mean, Median, Mode, Standard Deviation, Random variables, Discrete and Continuous distributions, Poisson distribution, Normal distribution, Binomial distribution, Correlation analysis, Regression analysis.
Section 2: Electric circuits

Network elements: ideal voltage and current sources, dependent sources, R, L, C, M elements; Network solution methods: KCL, KVL, Node and Mesh analysis; Network Theorems: Thevenin’s, Norton’s, Superposition and Maximum Power Transfer theorem; Transient response of dc and ac networks, sinusoidal steady-state analysis, resonance, two port networks, balanced three phase circuits, star-delta transformation, complex power and power factor in ac circuits.
Section 3: Electromagnetic Fields

Coulomb’s Law, Electric Field Intensity, Electric Flux Density, Gauss’s Law, Divergence, Electric field and potential due to point, line, plane and spherical charge distributions, Effect of dielectric medium, Capacitance of simple configurations, Biot‐Savart’s law, Ampere’s law, Curl, Faraday’s law, Lorentz force, Inductance, Magnetomotive force, Reluctance, Magnetic circuits, Self and Mutual inductance of simple configurations.
Section 4: Signals and Systems

Representation of continuous and discrete time signals, shifting and scaling properties, linear time invariant and
causal systems, Fourier series representation of continuous and discrete time periodic signals, sampling theorem,
Applications of Fourier Transform for continuous and discrete time signals, Laplace Transform and Z transform.
Section 5: Electrical Machines

Single phase transformer: equivalent circuit, phasor diagram, open circuit and short circuit tests, regulation and
efficiency;
Three-phase transformers: connections, vector groups, parallel operation; Auto-transformer, Electromechanical energy conversion principles;
DC machines: separately excited, series and shunt, motoring and generating mode of operation and their characteristics, speed control of dc motors;
Three-phase induction machines: principle of operation, types, performance, torque-speed characteristics, no-load and blocked-rotor tests, equivalent circuit, starting and speed control; Operating principle of single-phase induction motors;
Synchronous machines: cylindrical and salient pole machines, performance and characteristics, regulation and parallel operation of generators, starting of synchronous motors; Types of losses and efficiency calculations of electric machines
Section 6: Power Systems

Basic concepts of electrical power generation, ac and dc transmission concepts, Models and performance of transmission lines and cables, Series and shunt compensation, Electric field distribution and insulators, Distribution systems, Per‐unit quantities, Bus admittance matrix, Gauss- Seidel and Newton-Raphson load flow methods, Voltage and Frequency control, Power factor correction, Symmetrical components, Symmetrical and unsymmetrical fault analysis, Principles of over‐current, differential, directional and distance protection; Circuit breakers, System stability concepts, Equal area criterion, Economic Load Dispatch (with and without considering transmission losses).
Section 7: Control Systems

Mathematical modeling and representation of systems, Feedback principle, transfer function, Block diagrams and Signal flow graphs, Transient and Steady‐state analysis of linear time invariant systems, Stability analysis using Routh-Hurwitz and Nyquist criteria, Bode plots, Root loci, Lag, Lead and Lead‐Lag compensators; P, PI and PID controllers; State space model, Solution of state equations of LTI systems, R.M.S. value, average value calculation for any general periodic waveform.
Section 8: Electrical and Electronic Measurements

Bridges and Potentiometers, Measurement of voltage, current, power, energy and power factor; Instrument transformers, Digital voltmeters and multimeters, Phase, Time and Frequency measurement; Oscilloscopes, Error analysis.
Section 9: Analog Electronics and Digital Electronics

Simple diode circuits: clipping, clamping, rectifiers; Amplifiers: biasing, equivalent circuit and frequency response; oscillators and feedback amplifiers; operational amplifiers: characteristics and applications; single stage active filters, Sallen Key, Butterworth, VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers, Schmitt triggers, sample and hold circuits, A/D and D/A converters.
Section 10: Power Electronics

Static V-I characteristics and firing/gating circuits for Thyristor, MOSFET, IGBT; DC to DC conversion: Buck, Boost and Buck-Boost Converters; Single and three-phase configuration of uncontrolled rectifiers; Voltage and Current commutated Thyristor based converters; Bidirectional ac to dc voltage source converters; Magnitude and Phase of line current harmonics for uncontrolled and thyristor based converters; Power factor and Distortion Factor of ac to dc converters; Single-phase and three-phase voltage and current source inverters, sinusoidal pulse width modulation.

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GATE EE 2014 SET 3

Question 1
Two matrices A and B are given below:
A=\begin{bmatrix} p &q \\ r& s \end{bmatrix}; B=\begin{bmatrix} p^2+q^2 & pr +qs \\ pr+qs & r^2+s^2 \end{bmatrix}
If the rank of matrix A is N, then the rank of matrix B is
A
N/2
B
N-1
C
N
D
2N
Engineering Mathematics   Linear Algebra
Question 1 Explanation: 
\begin{aligned} A &=\begin{bmatrix} p & q\\ r & s \end{bmatrix} \\ A \times A=A^2&=\begin{bmatrix} p^2+q^2 & pr+qs\\ pr+qs & r^2+s^2 \end{bmatrix} =B \\ A^2 &=B \end{aligned}
Rank of amtrix does not change when we squaring the matrix, hence rank of B = rank of A=N.
Question 2
A particle, starting from origin at t=0s, is traveling along x-axis with velocity
v=\frac{\pi}{2}\cos (\frac{\pi}{2}t)m/s
At t=3s, the difference between the distance covered by the particle and the magnitude of displacement from the origin is_____
A
1
B
2
C
3
D
4
Engineering Mathematics   Calculus
Question 3
Let \triangledown \cdot (f v)=x^2y+y^2z+z^2x, where f and v are scalar and vector fields respectively. If v=yi+zj+xk , then v\cdot \triangledown f is
A
x^2y+y^2z+z^2x
B
2xy+2yz+2zx
C
x+y+z
D
0
Engineering Mathematics   Calculus
Question 3 Explanation: 
\begin{aligned} \vec{V}&=y\hat{i}+z\hat{j}+x\hat{k}\\ \hat{i}\frac{\partial (fV)}{\partial x}+\hat{j}\frac{\partial (fV)}{\partial y}+\hat{k}\frac{\partial (fV)}{\partial z}&=x^2y+y^2z+z^2x\\ y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}&=x^2y+y^2z+z^2x\;\;...(i)\\ \vec{V}\cdot \Delta f&=y\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial y}+x\frac{\partial f}{\partial z}\;\;...(ii)\\ \text{From equations (i) and (ii)}\\ \vec{V}\cdot \Delta f&=x^2y+y^2z+z^2x \end{aligned}
Question 4
Lifetime of an electric bulb is a random variable with density f(x)=kx^2, where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is _____
A
0.85
B
0.42
C
0.25
D
0.75
Engineering Mathematics   Probability and Statistics
Question 4 Explanation: 
Life time of an electric bulb with density
f(x)=Kx^2
If minimum and maximum lifetimes of bulb are 1 and 2 years respectively then
\begin{aligned} \int_{1}^{2}Kx^2dx &=1\\ \left.\begin{matrix} K\frac{x^3}{3} \end{matrix}\right|_1^2&=1\\ K\left ( \frac{8}{3}-\frac{1}{3} \right )&=1\\ \frac{7K}{3}&=1\\ K&=\frac{3}{7}=0.42 \end{aligned}
Question 5
A function f(t) is shown in the figure.

The Fourier transform F(\omega) of f(t) is
A
real and even function of w
B
real and odd function of w
C
imaginary and odd function of w
D
imaginary and even function of w
Signals and Systems   Fourier Transform
Question 5 Explanation: 
Fiven signal f(t) is an odd signal. Hence, F(\omega ) is imaginary and odd function of \omega .
Question 6
The line A to neutral voltage is 10 \angle 15^{\circ}V for a balanced three phase star connected load with phase sequence ABC . The voltage of line B with respect to line C is given by
A
10 \sqrt{3}\angle 105^{\circ} V
B
10 \angle 105^{\circ} V
C
10 \sqrt{3}\angle -75^{\circ} V
D
-10 \sqrt{3}\angle 90^{\circ} V
Electric Circuits   Three-Phase Circuits
Question 6 Explanation: 
Given,
V_{AN}=10\angle 15^{\circ}volt
As the system is balanced and phase sequence is ABC, therefore,
V_{AN}=10\angle 15^{\circ}
V_{BN}=10\angle 135^{\circ}V and V_{CN}=10\angle 255^{\circ}V
\therefore voltage of line w.r.t. line C is
V_{BC}=V_{BN}-V_{CN}
V_{BC}=10\angle 255^{\circ}-10\angle 135^{\circ}
V_{BC}=10\sqrt{3}\angle -75^{\circ} Volt
Question 7
A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux coming out of the concentric spherical surface of radius R(\gt r) is
A
4\pi \varepsilon _{0}r
B
4\pi \varepsilon _{0}r^{2}
C
4\pi \varepsilon _{0}R
D
4\pi \varepsilon _{0}R^{2}
Electromagnetic Fields   Electrostatic Fields
Question 8
The driving point impedance Z(s) for the circuit shown below is
A
\frac{s^{4}+3s^{2}+1}{s^{3}+2s}
B
\frac{s^{4}+3s^{2}+4}{s^{2}+2}
C
\frac{s^{2}+1}{s^{4}+s^{2}+1}
D
\frac{s^{3}+1}{s^{4}+s^{2}+1}
Electric Circuits   Two Port Network and Network Functions
Question 8 Explanation: 


Driving point impedance, Z(s) is ,
Z(s)=s+\left ( \frac{\left ( s+\frac{1}{s} \right )\times \frac{1}{s}}{s+\frac{1}{s}+\frac{1}{s}} \right )
\;\;=s+\left ( \frac{s^2+1}{s^2} \right )\times \frac{s}{s^2+2}
\;\;=s+\frac{s^2+1}{s(s^2+2)}
\;\;=\frac{s^2(s^2+2)+^2+1}{s^3+2s}
Z(s)=\frac{s^4+3s^2+1}{s^3+2s}
Question 9
A signal is represented by
x(t)=\left\{\begin{matrix} 1 &|t| \lt 1 \\ 0& |t|\gt 1 \end{matrix}\right.
The Fourier transform of the convolved signal y(t)= x(2t)* x(t/2) is
A
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})sin(2 \omega )
B
\frac{4}{\omega ^{2}}sin(\frac{\omega }{2})
C
\frac{4}{\omega ^{2}}sin(2 \omega )
D
\frac{4}{\omega ^{2}}sin^{2} \omega
Signals and Systems   Fourier Transform
Question 9 Explanation: 
Given signal can be drawn as

Therefore,
\begin{aligned} &x(t)\leftrightarrow X(\omega )=2Sa(\omega ) \\ &\text{Now, } x(t)\leftrightarrow X(\omega )\\ &\text{then by time scaling,} \\ &x(at)\leftrightarrow \frac{1}{|a|}X(\omega /a) \\ &\therefore \; x(2t)\leftrightarrow Sa \left ( \frac{\omega }{2} \right )\;\;...(i) \\ &x\left ( \frac{t}{2} \right ) \leftrightarrow 4Sa(2\omega )\;\;...(ii)\\ &\text{Now, }y(t)=x(2t) \times x(t/2) \end{aligned}
Convolution in time domain multiplication in frequency domain
\begin{aligned} Y(\omega )&=4Sa\left ( \frac{\omega }{2}\right ) Sa(2\omega ) \\ Y(\omega )&=\frac{4\sin \left ( \frac{\omega }{2}\right ) }{\left ( \frac{\omega }{2}\right ) } \frac{\sin (2\omega )}{2\omega }\\ Y(\omega )&=\frac{4}{\omega ^2}\sin \left ( \frac{\omega }{2}\right ) \sin (2\omega ) \end{aligned}
Question 10
For the signal
f(t) = 3 \sin 8 \pi t + 6 \sin 12 \pi t + \sin 14 \pi t,
the minimum sampling frequency (in Hz) satisfying the Nyquist criterion is _____.
A
7
B
14
C
18
D
9
Signals and Systems   Sampling
Question 10 Explanation: 
\begin{aligned} f_{m_1} &=4Hz \\ f_{m_2} &=6Hz \\ f_{m_3} &=7Hz \end{aligned}
Then minimum sampling frequency satisfying the nyquist criterion is 7*2=14Hz.
There are 10 questions to complete.

GATE EE 2015 SET 1

Question 1
A random variable X has probability density function f(x) as given below:

f(x)=\left\{\begin{matrix} a+bx & for \; 0 \lt x \lt 1\\ 0& otherwise \end{matrix}\right.

If the expected value E[X]=2/3, then Pr[X \lt 0.5] is _____________.
A
0.25
B
0.5
C
0.75
D
1
Engineering Mathematics   Probability and Statistics
Question 1 Explanation: 
\begin{aligned} f(x)&=\left\{\begin{matrix} a+bx & \text{for }0 \lt x \lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now given }E(X)&=2/3\\ \int_{0}^{1}xf(x)dx&=\frac{2}{3}\\ \int_{0}^{1}x(a+bx)dx&=\frac{2}{3}\\ a\left ( \frac{x^2}{2} \right )_0^1+b\left ( \frac{x^3}{3} \right )_0^1&=\frac{2}{3}\\ a\left ( \frac{1}{2} \right )+b\left ( \frac{1}{3} \right )&=\frac{2}{3}\\ 3a+2b&=4\;\;...(i)\\ \text{Now, }\int_{0}^{1}f(x)dx &=1\\ (\text{total probability} & \text{ is always equal to 1})\\ \int_{0}^{1}(a+bx)dx&=\left ( ax+\frac{bx^2}{2} \right )_0^1=1\\ a+\frac{b}{2}&=1\\ 2a+b&=2\;\;...(ii)\\ \text{Now solving } & \text{ (i) and (ii), we get}\\ a&=0,b=2\\ So,\; \;f(x)&=\left\{\begin{matrix} 2x & \text{for }0 \lt x lt 1\\ 0 & \text{otherwise} \end{matrix}\right.&\\ \text{Now we need}&\\ P(x \lt 0.5)&=\int_{0}^{0.5}2xdx=2\left ( \frac{x^2}{2} \right )_0^{0.5}\\ &=0.5^2-0^2=0.25 \end{aligned}
Question 2
If a continuous function f(x) does not have a root in the interval [a,b], then which one of the following statements is TRUE?
A
f(a)\cdot f(b)=0
B
f(a)\cdot f(b) \lt 0
C
f(a)\cdot f(b) \gt 0
D
f(a)/ f(b)\leq 0
Engineering Mathematics   Calculus
Question 2 Explanation: 
Intermediate value theorem states that if a function is continious and f(a) \cdot f(b) \lt 0, then surely there is a root in (a,b). The contrapositive of this theorem is that if a function is continious and has no root in (a,b) then surely f(a) \cdot f(b) \geq 0. But since it is given that there is no root in the closed interval [a,b] it means f(a) \cdot f(b) \neq 0.
So surely f(a) \cdot f(b) \gt 0 which is choise(C).
Question 3
If the sum of the diagonal elements of a 2x2 matrix is -6, then the maximum possible value of determinant of the matrix is ________.
A
6
B
8
C
9
D
12
Engineering Mathematics   Linear Algebra
Question 3 Explanation: 
Consider a symmetric matrix A =\begin{bmatrix} a & b\\ b & d \end{bmatrix}
Given a+d=-6
|A|=ad-b^2
Now since b^2 is always non-negative, maximum determinant will come when b^2=0.
So we need to maximize
\begin{aligned} |A|&=ad-0=ad=a \times -(6+a) \\ &= -a^2-6a\\ \frac{d|A|}{da} &=-2a-6=0 \\ \Rightarrow \;\; a &=-3 \text{ is the only stationary point} \end{aligned}
Since, \left [\frac{d^2|A|}{da} \right ]_{a=-3}=-2 \lt 0, we have a maximum at a=-3.
Since a+d=-6, corresponding value of d=-3.
|A|=ad=-3 \times -3=9
Question 4
Consider a function \vec{f}=\frac{1}{r^{2}}\hat{r}, where r is the distance from the origin and \hat{r} is the unit vector in the radial direction. The divergence of this function over a sphere of radius R, which includes the origin, is
A
0
B
2\pi
C
4\pi
D
R\pi
Electromagnetic Fields   Coordinate Systems and Vector Calculus
Question 4 Explanation: 
\bar{f}=\frac{1}{r^2}\hat{r}
From divergence theorem as we know,
\begin{aligned} \int _{vol.} (\triangledown \cdot \bar{f})dV &=\oint \oint _S \bar{f}\cdot d\bar{S}\\ \oint \oint _S \bar{f}\cdot d\bar{S} &= \oint \oint _S \left ( \frac{1}{r^2} \cdot \hat{r}\right ) \times r^2 \sin \theta \cdot d\theta \cdot d\phi \cdot \hat{r} \\ &= \oint \oint _S \sin \theta \cdot d\theta \cdot d\phi=4\pi \end{aligned}
Question 5
When the Wheatstone bridge shown in the figure is used to find the value of resistor R_X, the galvanometer G indicates zero current when R_1=50\Omega ,R_2=65\Omega and R_3=100\Omega. If R_3 is known with \pm 5% tolerance on its nominal value of 100 \Omega , what is the range of R_X in Ohms?
A
[123.50, 136.50]
B
[125.89, 134.12]
C
[117.00, 143.00]
D
[120.25, 139.75]
Electrical and Electronic Measurements   Characteristics of Instruments and Measurement Systems
Question 5 Explanation: 
R_1=50\Omega
R_2=60\Omega
R_3=100\pm 5%\Omega
The value of R_3 \text{with} \pm 5% of tollerance,
\begin{aligned} R_3&= 100\pm 5\% \\ &= 100+100 \times \frac{5}{100}=105 \Omega \\ &= 100-100 \times \frac{5}{100}=95 \Omega \end{aligned}
In both condition, the bridge is balance, so under balance condition,
\begin{aligned} \frac{R_1}{R_3}&=\frac{R_2}{R_x} \\ R_x&=\frac{R_2R_3}{R_1} \\ (i) \; \text{when}, \; R_3&=105\Omega \\ \therefore \;\;R_x &= \frac{105 \times 65}{50}=136.50\Omega \\ (ii)\; \text{when}, \; R_3&=95 \Omega \\ \therefore \;\;R_x &= \frac{95 \times 65}{50}=123.50\Omega \end{aligned}
Question 6
A (0-50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0-500 A) is _______.
A
0.11
B
0.22
C
0.45
D
0.68
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
Question 6 Explanation: 


I_{fs}=50A,
V_m=0.1V,
R_m=\frac{0.1}{50}=2 \times 10^{-3}\Omega
\because \; m=10
R_{sh}=\frac{R_m}{(m-1)}
\;\;\;=\frac{2 \times 10^{-3}}{9}=0.22\Omega
Question 7
Of the four characteristics given below, which are the major requirements for an instrumentation amplifier?

P. High common mode rejection ratio
Q. High input impedance
R. High linearity
S. High output impedance
A
P, Q and R only
B
P and R only
C
P, Q and S only
D
Q, R and S only
Analog Electronics   Operational Amplifiers
Question 8
In the following chopper, the duty ratio of switch S is 0.4. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is_____.
A
1
B
1.5
C
2.5
D
3
Power Electronics   Choppers
Question 8 Explanation: 


Step down chopper
\begin{aligned} \Rightarrow \; V_0&=\alpha V_s\\ V_0&=I_0R+E\\ I_0&=\frac{V_0-E}{R}\\ I_0&=\frac{\alpha V_s-E}{R}\\ &=\frac{(0.4)20-5}{3}=1A \end{aligned}
Question 9
A moving average function is given by y(t)=\frac{1}{T}\int_{t-T}^{t}u(\tau )d\tau. If the input u is a sinusoidal signal of frequency \frac{1}{2\tau } Hz, then in steady state, the output y will lag u (in degree) by ______ .
A
30
B
60
C
90
D
120
Signals and Systems   Linear Time Invariant Systems
Question 9 Explanation: 
System input: \sin \omega _0 t
\begin{aligned} f_0&=\frac{1}{2T}Hz\\ \text{Therefore,}\\ \omega _0&=2 \pi f_0=\frac{2 \pi}{2T}=\frac{\pi}{T}\; rad/sec\\ y(t)&=\frac{1}{T}\int_{t-T}^{t}\sin \omega _0\tau \; d\tau\\ &=\frac{1}{T}[-\cos\omega _0 t ]_{t-T}^t\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\omega _0 (t-T) ]\\ &\left [ \because \; \omega _0=\frac{t}{T} \right ]\\ &=-\frac{1}{T}[\cos\omega _0 t -\cos\left ( \frac{\pi t}{T}- \pi \right ) ]\\ \\ &=-\frac{1}{T}[\cos\omega _0 t +\cos\omega _0 t ]\\ &=-\frac{2}{T} \cos\omega _0 t \\ &=\frac{2}{T}\sin (\omega _0 t -90^{\circ}) \\ &=\text{Output is lagging by }90^{\circ} \text{ w.r.t. input.} \end{aligned}
Question 10
The impulse response g(t) of a system, G , is as shown in Figure (a). What is the maximum value attained by the impulse response of two cascaded blocks of G as shown in Figure (b)?
A
\frac{2}{3}
B
\frac{3}{4}
C
\frac{4}{5}
D
1
Signals and Systems   Linear Time Invariant Systems
Question 10 Explanation: 
\begin{aligned} g(t)&=u(t)-u(t-1)\\ G(s)&=\frac{1}{s}-\frac{e^{-s}}{s}\\ G(s) \times G(s)&=g(t)* g(t) \end{aligned}

Maximum value =1
There are 10 questions to complete.

GATE EE 2015 SET 2

Question 1
Given f(z)=g(z)+h(z), where f, g, h are complex valued functions of a complex variable z. Which one of the following statements is TRUE?
A
If f(z) is differentiable at z_0, then g(z) and g(z) are also differentiable at z_0.
B
If g(z) and h(z) are differentiable at z_0, then f(z) is also differentiable at z_0.
C
If f(z) is continuous at z_0, then it is differentiable at z_0.
D
If f(z) is differentiable at z_0, then so are its real and imaginary parts.
Engineering Mathematics   Complex Variables
Question 2
We have a set of 3 linear equations in 3 unknowns. 'X \equiv Y' means X and Y are equivalent statements and 'X\not\equiv Y' means X and Y are not equivalent statements.

P: There is a unique solution.
Q: The equations are linearly independent.
R: All eigenvalues of the coefficient matrix are nonzero.
S: The determinant of the coefficient matrix is nonzero.

Which one of the following is TRUE?
A
P\equiv Q\equiv R\equiv S
B
P\equiv R\not\equiv Q\equiv S
C
P\equiv Q\not\equiv R\equiv S
D
P\not\equiv Q\not\equiv R\not\equiv S
Engineering Mathematics   Linear Algebra
Question 2 Explanation: 
\begin{aligned} a_{11}x_1+a_{12}x_2+a_{13}x_3 &=b_1 \\ a_{21}x_1+a_{22}x_2+a_{23}x_3 &=b_2 \\ a_{31}x_1+a_{32}x_2+a_{33}x_3 &=b_3 \\ \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22}& a_{21}\\ a_{31}& a_{32} &a_{33} \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}&= \begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix} \end{aligned}
if |A|\neq 0 then AX=B can be written as X=A^{-1}B
It leads unique solutions.
If |A|\neq 0 then \lambda _1 \cdot \lambda _2\cdot \lambda _3\neq 0 each \lambda _i is non-zero.
If |A|\neq 0 then all the row (column) vectors of A are linearly independent.
Question 3
Match the following.
A
P-2 Q-1 R-4 S-3
B
P-4 Q-1 R-3 S-2
C
P-4 Q-3 R-1 S-2
D
P-3 Q-4 R-2 S-1
Electromagnetic Fields   Coordinate Systems and Vector Calculus
Question 3 Explanation: 
Stokes theorem \oint \vec{A}\cdot dl=\int \int (\triangledown \times A)\cdot \hat{n}ds
Gauss theorem \int \int D\cdot ds=Q
Divergence theorem \oint \oint A\cdot \hat{n}ds=\int \int \int \triangledown \cdot \bar{A}dV
Cauchy integral theorem \oint _cf(z)dz=0
Question 4
The Laplace transform of f(t)=2\sqrt{t/\pi } is s^{-3/2}. The Laplace transform of g(t)=\sqrt{1/\pi t} is
A
3s^{-5/2}/2
B
s^{-1/2}
C
s^{1/2}
D
s^{3/2}
Signals and Systems   Laplace Transform
Question 4 Explanation: 
Given that,
f(t)=2\sqrt{\frac{t}{\pi}}\rightleftharpoons F(s)=s^{-3/2}
By using property of differentiation In time,
\begin{aligned} \frac{df(t)}{dt}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{2}{\sqrt{\pi}}\cdot \frac{1}{2}t^{-1/2}&\rightleftharpoons sF(s) \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s\cdot s^{-3/2} \\ \Rightarrow \; \frac{1}{\sqrt{\pi t}}&\rightleftharpoons s^{-1/2} \end{aligned}
Question 5
Match the following.
A
P-1 Q-2 R-1 S-3
B
P-1 Q-2 R-1 S-3
C
P-1 Q-2 R-3 S-3
D
P-3 Q-1 R-2 S-1
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
Question 6
A 3-phase balanced load which has a power factor of 0.707 is connected to a balanced supply. The power consumed by the load is 5 kW. The power is measured by the two-wattmeter method. The readings of the two wattmeters are
A
3.94 kW and 1.06 kW
B
2.50 kW and 2.50 kW
C
5.00 kW and 0.00 kW
D
2.96 kW and 2.04 kW
Electrical and Electronic Measurements   Measurement of Energy and Power
Question 6 Explanation: 
\begin{aligned} p.f. &= 0.707\\ \phi &= cos^{-1}(0.707)=45^{\circ}\\ P&= 5kW\\ W_1 +W_2&= 5\;\;...(i)\\ \tan \phi &=\frac{\sqrt{3(W_1-W_2)}}{(W-1+W_2)} \\ (W-1+W_2) \times 1 &= \sqrt{3}(W_1-W_2) \\ W_1-W_2 &= \frac{5}{\sqrt{3}} \;\;...(ii)\\ \text{From equation} & \text{ (i) and (ii),} \\ W-1&= 3.94kW\\ W_2&= 1.06kW \end{aligned}
Question 7
A capacitive voltage divider is used to measure the bus voltage V_{bus} in a high-voltage 50 Hz AC system as shown in the figure. The measurement capacitors C_1 \; and \; C_2 have tolerances of \pm10% on their nominal capacitance values. If the bus voltage V_{bus} is 100 kV rms, the maximum rms output voltage V_{out} (in kV), considering the capacitor tolerances, is __________.
A
8.52
B
11.95
C
16.35
D
22.25
Electrical and Electronic Measurements   Galvanometers, Voltmeters and Ammeters
Question 7 Explanation: 


V_{BUS} \text{is} 100 kV_{rms}
C_1=1\mu F \pm 10\%
C_2=9\mu F \pm 10\%
To get maximum output voltage we need minimum C_2 and maximum C_1,
So, C_2=8.1\mu F and C_1=1.1\mu F
So, V_{out\;rms}=\left ( \frac{C_1}{C_1+C_2} \right )V_{BUS_{rms}}
\;\;\;=11.95kV
Question 8
In the following circuit, the input voltage V_{in} is 100sin(100\pit). For 100\piRC=50, the average voltage across R (in Volts) under steady-state is nearest to
A
100
B
31.8
C
200
D
63.6
Power Electronics   Phase Controlled Rectifiers
Question 8 Explanation: 
Given,
\begin{aligned} 100 \pi RC&=50\\ 100 \pi\cdot \frac{RC}{2}&=25\\ \omega \tau &=25 \end{aligned}

For one cycle \omega \tau =2 \pi \; rad=6.28
[\omega \tau =25] \gt \gt [\omega \tau =2 \pi \; rad=6.28]
Each capacitor voltage is approximately equal to
\begin{aligned} V_m&=100V \\ V_0&=2V_m\simeq 200V \end{aligned}
Question 9
Two semi-infinite dielectric regions are separated by a plane boundary at y = 0. The dielectric constants of region 1 (y\lt0) and region 2 (y\gt0) are 2 and 5, respectively. Region 1 has uniform electric field \vec{E}=3\hat{a}_{x}+4\hat{a}_{y}+2\hat{a}_{z}, where \hat{a}_{x},\hat{a}_{y} and \hat{a}_{z} are unit vectors along the x, y and z axes, respectively. The electric field in region 2 is
A
3\hat{a}_{x} + 1.6\hat{a}_{y} + 2\hat{a}_{z}
B
1.2\hat{a}_{x} + 4\hat{a}_{y} + 2\hat{a}_{z}
C
1.2 \hat{a}_{x} + 4\hat{a}_{y} + 0.8\hat{a}_{z}
D
3\hat{a}_{x} + 10\hat{a}_{y} + 0.8\hat{a}_{z}
Electromagnetic Fields   Electrostatic Fields
Question 9 Explanation: 
Given that,
at the interface (y=0) there is no surface charge.
Region 1 (y \lt 0) has \varepsilon _r=2
Region 2 (y \gt 0) has \varepsilon _r=5
Electric field in region 1 is 3a_x+4a_y+2a_z
Normal component of electric field is 4a_y tangential component of electric field is 3a_x+2a_y
Now, E_1=E_{1t}+E_{1n}
The tangentail component of E_1= tangentail component of E_2
E_{1t}=E_{2t}=3a_x+2a_z
\begin{aligned} \varepsilon _1 E_{1n}&=\varepsilon _2 E_{2n}\\ E_{2n}&=\frac{\varepsilon _1 \times 4a_y}{\varepsilon _2} \\ &= \frac{2 \times 4a_y}{5}=1.6a_y\\ E_2&=E_{2t}+E_{2n}\\ &=3a_x+1.6a_y+2a_z \end{aligned}
Question 10
A circular turn of radius 1 m revolves at 60 rpm about its diameter aligned with the x-axis as shown in the figure. The value of \mu _{0} is 4\pi \times 10^{-7} in SI unit. If a uniform magnetic field intensity \vec{H}=10^{7}\hat{z} A/m is applied, then the peak value of the induced voltage, V_{turn} ( in Volts), is _________.
A
150.35
B
200.25
C
248.05
D
300.54
Electromagnetic Fields   Time Varying Fields
Question 10 Explanation: 


The circular turn rotate with 60 rpm, let the angle made by ring w.r.t. x-axis \theta
and \;\;\; \theta =\omega _0 t
the turn rotate at 60 rpm,
so, \;\;\; \omega _0=2 \pi
So, the flux flowing through the circular turn wil be
\begin{aligned} \Psi &=(\mu _0 H_z \times \text{Area of turn} \times \cos \omega _0 t) \\ \Psi &= 4 \pi \times 10^{-7} \times 10^7\; A/m \times \pi \times 1^2 \times \cos \omega _0 t\\ &\text{Maximum voltage induced is } \\ \\ \left. \begin{matrix} \frac{d\Psi}{dt} \end{matrix}\right|_{max}&=(\omega _0 \times 4 \pi \times \pi \sin \omega _0 t)_{max}\\ V_{max}&=(4 \pi^2 \times 2 \pi)=248.05 volts \end{aligned}
There are 10 questions to complete.

GATE EE 2016 SET 1

Question 1
The maximum value attained by the function f(x) = x(x- 1)(x - 2) in the interval [1, 2] is _____.
A
0
B
1
C
2
D
4
Engineering Mathematics   Calculus
Question 1 Explanation: 
\begin{aligned} f(x) &=x^3-3x^2+2x \\ f'(x)&=3x^2-6x+2\\ f'(x)&=0 \;\text{for stationary point} \end{aligned}
stationary points are 1+\frac{1}{\sqrt{3}}
only 1+\frac{1}{\sqrt{3}} lies in [1,2]
\begin{aligned} f(1) &=0 \\ f(2)&=0 \\ f\left ( 1+\frac{1}{\sqrt{3}} \right )&=-\frac{2}{3\sqrt{3}} \end{aligned}
Maximum value is 0.
Question 2
Consider a 3 x 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is ____.
A
1
B
2
C
3
D
4
Engineering Mathematics   Linear Algebra
Question 2 Explanation: 
A=\begin{bmatrix} 1 & 1 &1 \\ 1 & 1 &1 \\ 1& 1 & 1 \end{bmatrix} Eigen value are 0,0,3
Question 3
The Laplace Transform of f(t)=e^{2t}sin(5t)u(t) is
A
\frac{5}{s^{2}-4s+29}
B
\frac{5}{s^{2}+5}
C
\frac{s-2}{s^{2}-4s+29}
D
\frac{5}{s+5}
Signals and Systems   Laplace Transform
Question 3 Explanation: 
Laplace transform of \sin 5t u(t)\rightarrow \frac{5}{s^2+25}
e^{2t}\sin 5t u(t)\rightarrow \frac{5}{(s-2)^2+25}=\frac{5}{s^2-4s+29}
Question 4
A function y(t), such that y(0)=1 and y(1)=3e^{-1}, is a solution of the differential equation \frac{d^{2}y}{dt^{2}}+2\frac{dy}{dt}+y=0. Then y(2) is
A
5e^{-1}
B
5e^{-2}
C
7e^{-1}
D
7e^{-2}
Engineering Mathematics   Calculus
Question 4 Explanation: 
Auxiliary equation,
\begin{aligned} m^2+2m+1&=0 \\ m&= -1,-1\\ y&=(c_1+c_2t)e^{-t} \\ y(0) &=1 \\ c_1&=1 \\ y&= (1+c_2t)e^{-t}\\ y(1)&=3e^{-1} \\ \Rightarrow \; (1+c_2)e^{-3} &=3e^{-3} \\ c_2&=2 \\ y&=(1+2t)e^{-t} \\ y(2)&=5e^{-2} \end{aligned}
Question 5
The value of the integral \oint_{c}\frac{2z+5}{(z-\frac{1}{2})(z^{2}-4z+5)}dz over the contour |z|=1, taken in the anti-clockwise direction, would be
A
\frac{24 \pi i}{13}
B
\frac{48 \pi i}{13}
C
\frac{24}{13}
D
\frac{12}{13}
Engineering Mathematics   Calculus
Question 5 Explanation: 
Singlarilies, Z=\frac{1}{2}, 2\pm i
Only, Z=\frac{1}{2} lies inside C
By residue theorem,
\oint _C=2 \pi i(R)=\frac{48 \pi i}{13}
Residue at \frac{1}{2}
=R_{1/2}=\lim_{Z \to 1/2}\left [ \left ( Z-\frac{1}{2} \right )\cdot \frac{2Z+5}{\left ( Z-\frac{1}{2} \right )(Z^2+4Z+5)} \right ]=\frac{24}{13}
Question 6
The transfer function of a system is \frac{Y(s)}{R(s)}=\frac{s}{s+2}. The steady state output y(t) is A \cos (2t+\varphi) for the input cos(2t). The values of A and \varphi, respectively are
A
\frac{1}{\sqrt{2}},-45^{\circ}
B
\frac{1}{\sqrt{2}},+45^{\circ}
C
\sqrt{2},-45^{\circ}
D
\sqrt{2},+45^{\circ}
Signals and Systems   Laplace Transform
Question 6 Explanation: 
\begin{aligned} \frac{Y(s)}{R(s)}&=\frac{s}{s+2} \\ y(t)&=A \cos (2t+\phi ), \\ r(t)&=\cos 2t \\ \because \;H(s) &=\frac{s}{(s+2)} \\ H(j\omega )&=\frac{j\omega }{j\omega +2} \\ |H(j\omega )| &=\frac{\omega }{\sqrt{\omega ^2+4 }} \\ \angle H(j\omega ) &=90^{\circ} -\tan ^{-1}\left ( \frac{\omega }{2} \right ) \\ \because \; \omega &= 2 \text{ (as given)}\\ |H(j\omega )| &=\frac{2}{\sqrt{4+4}}=\frac{1}{\sqrt{2}} \\ |H(j\omega )| &=90^{\circ} -\tan ^{-1}(1)=45^{\circ} \\ \because \; \text{hence, }A &=1 \times |H(j\omega )|_{\omega =2}\\ &=1 \times \frac{1}{\sqrt{2}}=0.707\\ \phi &= 45^{\circ} \end{aligned}
Question 7
The phase cross-over frequency of the transfer function G(s)=\frac{100}{(s+3)^{3}} in rad/s is
A
\sqrt{3}
B
1/\sqrt{3}
C
3
D
3\sqrt{3}
Control Systems   Frequency Response Analysis
Question 7 Explanation: 
G(s)=\frac{100}{(s+1)^3}
G(j\omega )=\frac{100}{(1+j\omega )^3}
\;\;=\frac{100}{1+(j\omega )^3+3(j\omega )^2+3j\omega }
\;\;=\frac{100}{(1-3\omega^2 )+j(3\omega-\omega^3)}
\;\;=\frac{100[(1-3\omega^2 )+j\omega(3-\omega^2)^2]}{[ (1-3\omega^2 )+\omega^2(3-\omega^2)^2]}
For phase corssover frequency \omega_{ph} Img[G(j \omega )]=0;
Hence, \omega (3-\omega ^2)=0
\omega =0; \pm \sqrt{3}
Therefore, \omega _{ph}=\sqrt{3} rad/sec
Question 8
Consider a continuous-time system with input x(t) and output y(t) given by

y(t) = x(t) cos(t)

This system is
A
linear and time-invariant
B
non-linear and time-invariant
C
linear and time-varying
D
non-linear and time-varying
Signals and Systems   Linear Time Invariant Systems
Question 8 Explanation: 
\begin{aligned} y(t)&=x(t)\cos (t)\\ &\text{To check linearity,}\\ y_1(t)&=x_1(t)\cos (t)\\ &[y_1(t) \text{ is output for }x_1(t)]\\ y_2(t)&=x_2(t) \cos (t)\\ &[y_2(t) \text{ is output for }x_2(t)]\\ \text{so, the}& \text{ output for }(x_1(t)+ x_2(t)) \text{ will be}\\ y(t)&=[x_1(t)+ x_2(t)]\cos (t)\\ &=y_1(t)+y_2(t) \end{aligned}
So, the system is linear, to check time invariance.
The delayed output,
y(t-t_0)=x(t-t_0)\cos (t-t_0)
The output for delayed input,
y(t, t_0)=x(t-t_0)\cos (t)
Since, y(t-t_0)\neq y(t,t_0)
System is time varying.
Question 9
The value of \int_{-\infty }^{+\infty }e^{-t}\delta (2t-2)dt, \; where \; \delta (t) is the Dirac delta function, is
A
\frac{1}{2e}
B
\frac{2}{e}
C
\frac{1}{e^{2}}
D
\frac{1}{2e^{2}}
Signals and Systems   Introduction of C.T. and D.T. Signals
Question 9 Explanation: 
To find the value of \int_{-\infty }^{\infty }e^{-t}\delta (2t-2)dt
Since, \delta (2t-2)=\frac{1}{2}\delta (1t-1) above integral can be written as
\int_{-\infty }^{\infty }e^{-t}\frac{1}{2}\delta (t-1)dt=\frac{1}{2}e^{-1}=\frac{1}{2e}
Question 10
A temperature in the range of -40^{\circ}C to 55^{\circ}C is to be measured with a resolution of 0.1^{\circ}C. The minimum number of ADC bits required to get a matching dynamic range of the temperature sensor is
A
8
B
10
C
12
D
14
Digital Electronics   A-D and D-A Converters
Question 10 Explanation: 
Temperature range of -40^{\circ}C \; to \; 55^{\circ}C
So. Total range in 95^{\circ}C
Since, resolution is 0.1^{\circ}C
So, number of steps will be 950
To have 950 steps, we need at least 10 bits.
There are 10 questions to complete.

GATE EE 2016 SET 2

Question 1
The output expression for the Karnaugh map shown below is
A
A+\bar{B}
B
A+\bar{C}
C
\bar{A}+\bar{C}
D
\bar{A}+C
Digital Electronics   Boolean Algebra and Minimization
Question 1 Explanation: 


F=A+\bar{C}
Question 2
The circuit shown below is an example of a
A
low pass filter.
B
band pass filter
C
high pass filter.
D
notch filter.
Analog Electronics   Operational Amplifiers
Question 2 Explanation: 


\frac{V_{out}}{V_{in}}=-\left [\frac{\frac{ R_2\cdot \frac{1}{j\omega C}}{R_2+\frac{1}{j\omega C}}}{R_1} \right ]
\frac{V_{out}}{V_{in}}=-\left [ \frac{R_2}{R_1(R_2j\omega C+1)} \right ]
So the system is a low pass filter.
Question 3
The following figure shows the connection of an ideal transformer with primary to secondary turns ratio of 1:100. The applied primary voltage is 100 V (rms), 50 Hz, AC. The rms value of the current I, in ampere, is __________.
A
100
B
10
C
20
D
50
Electric Circuits   Magnetically Coupled Circuits, Network Topology and Filters
Question 3 Explanation: 
The above circuit can be drown by transferring secondary circuit to primary side.

I=\frac{100V}{(8+10j-4j)\Omega }
\;\;=\frac{100V}{(8+6j)\Omega }
So the rms value of I will be 10 A.
Question 4
Consider a causal LTI system characterized by differential equation \frac{dy(t)}{dt}+\frac{1}{6}y(t)=3x(t). The response of the system to the input x(t)=3e^{-\frac{t}{3}}u(t). where u(t) denotes the unit step function, is
A
9e^{-\frac{t}{3}}u(t)
B
9e^{-\frac{t}{6}}u(t)
C
9e^{-\frac{t}{3}}u(t)-6e^{-\frac{t}{6}}u(t)
D
54e^{-\frac{t}{6}}u(t)-54e^{-\frac{t}{3}}u(t)
Signals and Systems   Linear Time Invariant Systems
Question 4 Explanation: 
The differential equation
\begin{aligned} \frac{dy(t)}{dt} &+\frac{1}{6}y(t)=3x(t) \\ \text{So, }sY(s)&+\frac{1}{6}Y(s) =3X(s) \\ Y(s) &=\frac{3X(s)}{\left ( s+\frac{1}{6} \right )} \\ X(s) &=\frac{9}{\left ( s+\frac{1}{3} \right )} \\ \text{So, } Y(s)&=\frac{9}{\left ( s+\frac{1}{3} \right )\left ( s+\frac{1}{6} \right )} \\ &=\frac{54}{\left ( s+\frac{1}{6} \right )} -\frac{54}{\left ( s+\frac{1}{3} \right )}\\ \text{So, }y(t) &= (54e^{-1/6t}-54e^{-1/3t})u(t) \end{aligned}
Question 5
Suppose the maximum frequency in a band-limited signal x(t) is 5 kHz. Then, the maximum frequency in x(t)\cos (2000\pi t), in kHz, is ________.
A
5
B
6
C
7
D
8
Signals and Systems   Fourier Transform
Question 5 Explanation: 
Maximum possible frequency of x(t)(2000 \pi t)=f_1+f_2=5+1=6kHz
Question 6
Consider the function f(z)=z+z^* where z is a complex variable and z^* denotes its complex conjugate. Which one of the following is TRUE?
A
f(z) is both continuous and analytic
B
f(z) is continuous but not analytic
C
f(z) is not continuous but is analytic
D
f(z) is neither continuous nor analytic
Engineering Mathematics   Complex Variables
Question 6 Explanation: 
\begin{aligned} f(z) &=z+z^* \\ f(z)&=2x \text{ is continuous (polynomial)} \\ u &=2x, v=0 \\ u_x &=2, u_y=0 \\ v_x&=0, v_y=0 \end{aligned}
C.R. equation not satisfied.
Therefore, no where analytic.
Question 7
A 3 x 3 matrix P is such that, P^{3} = P. Then the eigenvalues of P are
A
1, 1, -1
B
1, 0.5 + j0.866, 0.5 - j0.866
C
1, -0.5 + j0.866, -0.5 - j0.866
D
0, 1, -1
Engineering Mathematics   Linear Algebra
Question 7 Explanation: 
By Calyey Hamilton theorem,
\lambda ^3=\lambda
\lambda =0, 1, -1
Question 8
The solution of the differential equation, for t \gt 0, y''(t)+2y'(t)+y(t)=0 with initial conditions y(0) = 0 and y'(0) = 1, is (u(t) denotes the unit step function),
A
te^{-t}u(t)
B
(e^{-t}-te^{-t})u(t)
C
(-e^{-t}+te^{-t})u(t)
D
e^{-t}u(t)
Engineering Mathematics   Differential Equations
Question 8 Explanation: 
The differentail equation is
y'(t)+2y'(t)+y(t)=0
So, (s^2y(s)-sy(0)-y'(0))+2[sy(s)-y(0)]+y(s) =0
So, y(s)=\frac{sy(0)+y'(0)+2y(0)}{s^2+2s+1}
Given that y'(0)=1, y(0)=0
So, y(s)=\frac{1}{(s+1)^2}
So, y(t)=te^{-t}u(t)
Question 9
The value of the line integral
\int_{c}(2xy^{2}dx+2x^{2}ydy+dz)
along a path joining the origin (0,0,0) and the point (1,1,1) is
A
0
B
2
C
4
D
6
Engineering Mathematics   Calculus
Question 9 Explanation: 
\int _C\bar{F}\cdot \bar{dr}
where,
\bar{F}=xy^2\bar{i}+2x^2y\bar{j}+\bar{k}
\bigtriangledown \times \vec{F}=\vec{O}
(\vec{F} is irrotational \Rightarrow \vec{F} is conservative)
\begin{aligned} \vec{F}&=\bigtriangledown \phi \\ \phi _x&=2xy^2 \\ \phi _y&=2x^2y \\ \phi _z&= 1\\ \Rightarrow \; \phi &=x^2y^2+z+C \end{aligned}
where, \vec{F} is conservative
\int _C \bar{F}\bar{dr}=\int_{(0,0,0)}^{(1,1,1)}d\phi =[x^2y^2+z]_{(0,0,0)}^{(1,1,1,)}=2
Question 10
Let f(x) be a real, periodic function satisfying f(-x)=-f(x). The general form of its Fourier series representation would be
A
f(x)=a_{0}+\sum_{k=1}^{\infty }a_{k}cos(kx)
B
f(x)=\sum_{k=1}^{\infty }b_{k}sin(kx)
C
f(x)=a_{0}+\sum_{k=1}^{\infty }a_{2k}cos(kx)
D
f(x)=\sum_{k=0}^{\infty }a_{2k+1}sin(2k+1)x
Signals and Systems   Fourier Series
Question 10 Explanation: 
Given that,
f(-x)=-f(x)
So, function is an odd function.
So, the fourier series will have sine term only. So,
f(x)=\sum_{k=1}^{\infty }b_x \sin (kx)
There are 10 questions to complete.